Question: Evaluate $$\lceil\sqrt{5}\rceil + \lceil\sqrt{6}\rceil + \lceil\sqrt{7}\rceil + \cdots + \lceil\sqrt{29}\rceil$$Note: For a real number $x,$ $\lceil x \rceil$ denotes the smallest integer that is greater than or equal to $x.$
We note that if $a^2 < n \leq (a+1)^2$ for some integer $a$, then $a < \sqrt{x} \leq a+1$, so $a$ is the least integer greater than or equal to $x$. Consequently, we break up our sum into the blocks of integers between consecutive perfect squares:

For $5\leq n \leq 9$, $\lceil\sqrt{n}\rceil=3$. There are $5$ values of $3$ in this range.

For $10\leq n\leq 16$, $\lceil\sqrt{n}\rceil=4$. There are $7$ values of $4$ in this range.

For $17\leq n \leq 25$, $\lceil\sqrt{n}\rceil=5$. There are $9$ values of $5$ in this range.

For $26\leq n \leq 29$, $\lceil\sqrt{n}\rceil=6$. There are $4$ values of $6$ in this range.

Consequently, our total sum is $5\cdot3+7\cdot4+9\cdot5+4\cdot 6= \boxed{112}$.